Talk:Emissivity

Old

The thermal emissivity page wasn't doing much, so I patched together material from other articles which linked directly to emissivity (which means that that article isn't doing much, either, but at least it's all in one place). As far as I know, it's all thermal emissivity. --Aranel 21:49, 15 Aug 2004 (UTC)

$\epsilon$ or $e$ ?

I wrote the original article on emissivity, on the portuguese Wikipedia. I refer to emissivity using the greek letter epsilon ( $\epsilon$ ), instead of the mathmatical format of e ( $e$ ). So, anyone has any idea on which one works best? --Arthur Albano 13:26, 18 July 2006 (UTC)[reply]

I've been trying to find out what is most common and my physics textbooks use

e

. They are pretty elementary though. I have found some mention of

\epsilon

on the web though.
—Apis (talk) 19:45, 28 January 2009 (UTC)[reply]

Infrared Emissivity

A search for 'infrared emissivity' returns 0 results in Wikipedia. If somebody is aware they could add, as the term returns many results in Google - Sridhar Nov 12, 2005

It's a narrow yet vague concept. Emissivity is defined independently for each wavelength over the whole electromagnetic spectrum. On the one hand the infrared is just a tiny part of the spectrum, hence narrow, on the other it nevertheless embraces a continuum of wavelengths ranging over more than an order of magnitude, hence vague. --Vaughan Pratt (talk) 06:57, 13 July 2010 (UTC)[reply]

2 Walls

Anyone know where the information on the emissivity of 2 walls comes from? Joseph449008 (talk) 14:00, 29 January 2010 (UTC)[reply]

No idea, but it's easy to see that it's true. The formula gives the proportion of radiation (of whatever wavelength the two emissivities are for) inside one wall that enters the other wall. It takes into account Kirchhoff's law of thermal radiation, that reflectivity is complementary to emissivity. The reflectivity allows computing the amount of radiation bouncing back and forth forever in between the two walls as a result of reflection. I'll try to say something coherent about this in the article.

To find the formula, assume unit radiation inside the first wall, and write R1 = 1-E1, R2 = 1-E2 for the two reflectivities. Let A be the radiation from the first wall to the second, and B the radiation the other way. Then A = E1 + R1*B (sum of emitted and reflected radiation) while B = R2*A (just the reflected part of A). Eliminating B gives A = E1/(1-R1*R2). What is emitted into the second wall is then E2*A = E1*E2/(1-R1*R2) and it's now just algebra to verify that this is the formula in the article (expand R1 and R2, simplify, then divide both top and bottom by E1*E2). The formula is symmetric so it gives the same result for radiation from inside the first wall to inside the second. Note that "inside" means "just inside" so that absorption does not become an issue. --Vaughan Pratt (talk) 05:17, 13 July 2010 (UTC)[reply]

albedo

relation to albedo..? Cesiumfrog (talk) 07:36, 27 April 2010 (UTC)[reply]

By Kirchhoff's law of thermal radiation for a given wavelength, emissivity is simply one minus reflectivity. Naively albedo is just reflectivity, but in practice it's much more complicated and there's no simple connection. While there are various notions of albedo, for example geometric albedo and Bond albedo, none of them are for a single wavelength, whence there is no notion of albedo for which Kirchhoff's law can be trusted.

For a body whose reflectivity is independent of wavelength, Kirchhoff's law does hold, but in all other situations it's at best a rough approximation. If this were not so the formula for effective temperature, which assumes unit emissivity and therefore underestimates the temperature of a planet with no atmosphere, could solve this very simply by just dropping the factor 1-A, which would then factor in the emissivity. --Vaughan Pratt (talk) 06:50, 13 July 2010 (UTC)[reply]

The Definition is Wrong !

The word 'power' should be substituted for 'energy' throughout. Another Wikki cock-up! A.J Smith

How do you figure? What gets radiated is energy. Power is how much per unit time. If the usage is wrong some place, identify the place for us, or fix it. Dicklyon (talk) 22:47, 6 March 2011 (UTC)[reply]

Astrophysical Greybodies?

Shouldn't $\beta =1$ for dust clouds in space? Relationship between frequency and extinction is based on Mie Theory, which $\sim 1/\lambda$ . Is this not correct? 128.146.32.165 (talk) 18:57, 5 March 2012 (UTC) Paul Rimmer[reply]

color vs emissivity

The rev 23:45, 19 May 2008 added a second paragraph that was probably meant as a quick aid to understanding of the definition:

"In general, the duller and blacker a material is, the closer its emissivity is to 1. The more reflective a material is, the lower its emissivity. Highly polished silver has an emissivity of about 0.02."

Unfortunately it introduced some problems. It included a use of the term "blacker" that both depended on the definition of emissivity itself (regarding black bodies) and greatly tempted a misconception about how color (not similarities to a black body) relates to emissivity. Also it generalized metallic properties across all materials which was misleading, and it failed to mention a more primary influencial factor--the metallic vs nonmetallic nature of a material.

Good intent by the author, but it even looks like other sites have copied the sentence verbatim into their explanations of emissivity also. I've reworked it for now to make it more clear, but it could use some more work when someone gets time. If there's benefit at all to having a sentence or two like that after the definition, the challenge is probably going to be expressing something of value that's short and concise enough to defer details until the explanation section below it.

Senortres (talk) 22:50, 11 January 2014 (UTC)[reply]

Original research: section on Emissivity between two walls

This article must not contain original research. The Emissivity between two walls section automatically assumes that the fraction of light of any frequency emitted compared to that of a black-body is equal to the fraction of that frequency that it would absorb. That information though not explixitly stated was assumed in a formula and is not backed up by any of this article's references. Blackbombchu (talk) 20:38, 19 March 2014 (UTC)[reply]

I think this section is only incidentally related to the article's subject. Emissivity is a property of a single surface; I have never seen the term applied to "two walls". I think the section is original research, as you suggest. I'll delete this section within the next day or so unless some one provides a reference to this usage. Easchiff (talk) 10:35, 19 July 2014 (UTC)[reply]

It was one of many topics in the old page that would have probably been better separated out or referenced out to a separate page anyway, it seemed to be overly complicating the focus with emission interactions. By the way, great work on the updates to the page E, it's looking a ton better. Senortres (talk) 03:00, 14 August 2014 (UTC)[reply]

Thanks very much for the compliment - always good to know when WikiWork is appreciated! Best, Easchiff (talk) 03:06, 14 August 2014 (UTC)[reply]

Anodized aluminum: effect of color?

On the section: "Emissivities of common surfaces", what color is the anodization in 'Aluminum, anodized'? There should be two emissivities listed; one for black, one for clear. 12.33.223.211 (talk) 20:05, 5 May 2016 (UTC)[reply]

Possibly not. The properties at wavelengths near 10,000 nm, which are the ones that determine emissivity, are profoundly different than properties at much shorter wavelengths that are visible (400-700 nm). Do you have a citation indicating that it matters? Cheers, Easchiff (talk) 22:42, 5 May 2016 (UTC)[reply]

It's a good question. The visible color of the anodized surface has little effect. I've added a citation in the article. The probable reason, in my (unpublished) opinion: aluminum anodizing typically produces an aluminum oxide coating. The coating has a high emissivity, meaning that it is quite "black" at the wavelengths that matter for emissivity. However, aluminum oxide is nearly clear at visible wavelengths; clear sapphire gemstones are pure aluminum oxide crystals. So the inorganic dyes that are used to give a visible color are noticeable because the aluminum oxide is clear at those wavelengths. If the anodized coating were already black at visible wavelengths, one wouldn't expect dying to affect the color much. This is what happens at the much longer wavelengths that matter for emissivity. Easchiff (talk) 05:36, 7 May 2016 (UTC)[reply]

Nice response; I read the citation. Surprising result. Thank you! 71.139.160.208 (talk) 19:46, 7 May 2016 (UTC)[reply]

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Is emissivity complementary to reflectivity?

My understanding is that (ignoring wavelength-dependent differences) the more reflective an object is, the less emissive it is, and vice-versa. If this statement is indeed valid (even with caveats), it would be great to add to the article, as it would make it much easier for the casual reader to understand what emissivity is. -- Dan Griscom (talk) 23:03, 7 January 2018 (UTC)[reply]

As well as reflectivity, there's also transmissivity. Incident energy that is not absorbed (emissivity = absorptivity; https://en.wikipedia.org/wiki/Kirchhoff%27s_law_of_thermal_radiation) is either reflected or transmitted through. A good example is that glass transmits solar thermal radiation wavelengths (~500nm) efficiently, but has a reflectivity spectrum peak at room temperature wavelengths (~10um). This article is about emissivity though. Gideonr0 (talk) 14:06, 27 November 2024 (UTC)[reply]

(Not an) erroneous calculation?!

I was about to challenge the quote: "The surface of a perfect black body (with an emissivity of 1) emits thermal radiation at the rate of approximately 448 watts per square metre at room temperature," but realize that I was assuming they meant W/cm2. They clearly state W/m2, so that seems much more believable. 66.219.244.122 (talk) 21:31, 20 December 2021 (UTC)MCG[reply]

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Old

ϵ {\displaystyle \epsilon } or e {\displaystyle e} ?